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Question 9.4: Given the function (a) Sketch the function. (b) Determine it...

Given the function

f(x)={0, for            −5≤x≤0,3,for                    0≤x≤5f(x)=\begin{cases} 0, &  for                         −5 ≤ x ≤ 0,\\ 3, & for                     0 ≤ x ≤ 5 \end{cases}                    period 2l = 10

(a) Sketch the function.
(b) Determine its Fourier series.

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(a)

f(x)={0, for            −5≤x≤0,3,for                     0≤x≤5f(x)=\begin{cases} 0, &  for                         −5 ≤ x ≤ 0,\\ 3, & for                      0 ≤ x ≤ 5 \end{cases}                    period 2l = 10

(b) For period 2l = 10 and l = 5, we choose the interval a to a + 2l to be −5 to 5, i.e., a =−5:

an=1l∫aa+2lf(x) cosnπxldx=15∫−55f(x) cosnπxldxa_{n} = \frac{1}{l} \int\limits_{a}^{a+2l}{f (x)  cos \frac{nπx}{l} dx} = \frac{1}{5} \int\limits_{-5}^{5}{f (x)  cos \frac{nπx}{l}dx}

 

=15{∫−50(0) cosnπx5dx+∫053 cosnπx5dx}=35∫05cosnπx5dx= \frac{1}{5} \left\{\int\limits_{-5}^{0}{(0)  cos \frac{nπx}{5} dx} + \int\limits_{0}^{5}{3  cos \frac{nπx}{5} dx} \right\}= \frac{3}{5} \int\limits_{0}^{5}{cos \frac{nπx}{5}dx}

 

=35{5nπsinnπx5}∣05=0      for      n≠0.= \frac{3}{5} \left\{\frac{5}{nπ} sin \frac{nπx}{5}\right\}|^{5}_{0}=0       for             n ≠ 0.

For n = 0, one has an=a0=(3/5)∫05cos(0πx/5)dx=(3/5) ∫05dx=3.a_{n} = a_{0} = (3/5) \int_{0}^{5}{cos(0πx/5)dx} = (3/5)  \int_{0}^{5}{ dx} = 3.
Furthermore,

bn=1l∫aa+2lf(x) sinnπxldx=15∫−55f(x) sinnπxldxb_{n} = \frac{1}{l} \int\limits_{a}^{a+2l}{ f (x)  sin \frac{nπx}{l} dx} = \frac{1}{5} \int\limits_{-5}^{5}{ f (x)  sin \frac{nπx}{l} dx}

 

=15{∫−50(0) sinnπx5dx+∫053 sinnπx5dx}=35∫05sinnπx5dx= \frac{1}{5} \left\{\int\limits_{-5}^{0}{(0)  sin \frac{nπx}{5} dx} + \int\limits_{0}^{5}{3  sin \frac{nπx}{5} dx} \right\}= \frac{3}{5} \int\limits_{0}^{5}{sin \frac{nπx}{5}dx}

 

=35(−5nπcosnπx5)∣05=3nπ(1−cosnπ).= \frac{3}{5} \left(− \frac{5}{nπ} cos \frac{nπx}{5}\right)|^{5}_{0} = \frac{3}{nπ} (1 −cos nπ).

Thus,

f(x)=32+∑n=1∞3nπ(1−cosnπ)sin(nπx5),f (x) = \frac{3}{2}+ \sum\limits_{n=1}^{∞}{\frac{3}{nπ} (1 − cos nπ) sin \left(\frac{nπx}{5}\right)},

i.e.,

f(x)=32+6π(sinπx5+13sin3πx5+15sin5πx5+⋅⋅⋅).f (x) = \frac{3}{2}+ \frac{6}{π} \left(sin \frac{πx}{5}+ \frac{1}{3}sin \frac{3πx}{5} + \frac{1}{5}sin \frac{5πx}{5}+· · ·\right).
9.3

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